Variants of Turing Machines Question Computability and Complexity - PDF document

Variants of Turing Machines Question Computability and Complexity What happens if we modify the definition of a Turing machine? Can we then possibly recognize more languages? Lecture 2 Example: δ : Q × Γ → Q × Γ × { L , R , S } where S means “stay”. Multitape Turing machines Nondeterministic Turing machines Answer Enumerators Church-Turing Thesis No, the notion of a TM is robust. Hence no reasonable extension of a TM increases its power. given by Jiri Srba Example: “Stay” can be simulated in ordinary TM by two head movements (move right, move left). Lecture 2 Computability and Complexity 1/15 Lecture 2 Computability and Complexity 2/15 Multitape Turing Machine Equivalence of 1-tape and k -tape TM Definition (Multitape Turing Machine) Theorem A k -tape TM is a 7-tuple M = ( Q , Σ , Γ , δ, q 0 , q accept , q reject ) where Every k -tape TM is equivalent to 1-tape TM. δ : Q × Γ k → Q × Γ k × { L , R } k Equivalent means that they recognize the same language. and the rest is the same as before. Remark: 1-tape TM is exactly our original definition of TM. Corollary A language is recognizable iff it is recognized by some multitape Turing machine. Configuration: a control state, plus the content of all k tapes together with the position of k heads. Corollary Initial configuration: input string written on the first tape, all other tapes are empty (contain the blank symbols). A language is decidable iff it is recognized by some multitape Turing machine which is a decider. Computational step: all heads can move independently. Lecture 2 Computability and Complexity 3/15 Lecture 2 Computability and Complexity 4/15 Nondeterministic Turing Machine Equivalence of Deterministic and Nondeterministic TM Theorem Definition (Nondeterministic Turing Machine) Every nondeterministic TM is equivalent to some deterministic TM. A nondeterministic TM is a 7-tuple M = ( Q , Σ , Γ , δ, q 0 , q accept , q reject ) where Equivalent means that they recognize the same language. δ : Q × Γ → 2 Q × Γ ×{ L , R } Corollary and the rest is the same as before. A language is recognizable iff it is recognized by some nondeterministic Turing machine. Remark: Every deterministic TM is also a nondeterministic TM. Configurations and Initial configuration as before. Corollary Computation tree: a tree of all configurations reachable from A language is decidable iff it is recognized by some the initial one. The tree can have infinite branches! nondeterministic Turing machine which is a decider. Acceptance Condition of a Nondeterm. TM A nondeterministic TM M accepts a string w if the computation A nondeterministic TM is a decider if for any given input every tree for M and w contains at least one accepting configuration. branch in the computation tree is finite (accepts or rejects). Lecture 2 Computability and Complexity 5/15 Lecture 2 Computability and Complexity 6/15

Enumerators Theorem about Enumerators Another terminology for recognizable languages is the term Recursively Enumerable languages. Why? Definition (Enumerator) An enumerator is a 2-tape Turing machine with a special control Theorem state called q print . A language L is recognizable if and only if there exists an enumerator E that enumerates L , i.e. L ( E ) = L . Definition (Language Generated by Enumerator) Let E be an enumerator. We run E on the empty string as input. The language of E , denoted by L ( E ), is the collection of all strings that are on the second tape whenever E is in the state q print . Remark: If E does not terminate then L ( E ) may be infinite. Strings in L ( E ) may repeat and may be printed in arbitrary order. Lecture 2 Computability and Complexity 7/15 Lecture 2 Computability and Complexity 8/15 Proof (Enumerable = ⇒ Recognizable) Proof (Recognizable = ⇒ Enumerable) Every recognizable language L is enumerable: Let M be a recognizer for L . We construct an enumerator E for L . Every enumerable language L is recognizable: Let s 1 , s 2 , s 3 , . . . be all possible strings from Σ ∗ . Let E be an enumerator for L . We construct a recognizer M for L . E = ” 1. i := 1; M = ” On input w : 2. Run M for i steps on each input s 1 , s 2 , . . . , s i . 1. Run E . 3. If M accepted any of the strings, print it 2. If w gets ever printed then M accepts, 4. i := i +1; goto step 2.” otherwise continue running E in step 1.” This technique is called Dovetailing. Lecture 2 Computability and Complexity 9/15 Lecture 2 Computability and Complexity 10/15 Hilbert’s Tenth Problem Church-Turing Thesis Models of an algorithm: 1936: Alan Turing came with Turing machine and Alonzo In 1900 David Hilbert asked to find a mechanical way to Church with λ -calculus. check whether a polynomial (over several variables) has an Turing machines and λ -calculus were shown equivalent. integral root. Many more models suggested: Kleene’s µ -recursive functions, Example: Post-systems, Minsky machine ... all shown equivalent! 6 x 3 yz 2 + 3 xy 2 − x 3 − 10 Church-Turing Thesis has an integral root at x = 5, y = 3 and z = 0. Algorithms = Turing machines Answer (informal notion) (formal, mathematical, concept) Nobody has found such an algorithm yet ... in fact, we know that Facts: this is impossible, and we can prove this!!! (Yuri Matijasevic’1970). Church-Turing Thesis cannot be proved, BUT ... We need a model of an algorithm to demonstrate such a proof. any C/Java/C++/C# program can be run on a TM, and nothing better than a Turing machine has been found so far. Lecture 2 Computability and Complexity 11/15 Lecture 2 Computability and Complexity 12/15

Algorithmic Problems vs. Decidable Languages Algorithmic Problems vs. Decidable Languages: Examples How do algorithmic problems correspond to languages? Graph Connectivity “Is a given graph G connected?” corresponds to: Example: Hilbert’s Tenth Problem Does � G � (encoding of G ) belong to the language Can the problem whether a given polynomial has an integral root def = {� G ′ � | G ′ is a graph and G ′ is connected } ? be algorithmically solved? L connected Language Formulation of Hilbert’s Tenth Problem Acceptance Problem of a TM “Does a given TM M accept a string w ?” corresponds to: D def = {� p � | p is a polynomial with integral root } where � p � is the textual (string) encoding of the polynomial p . Does � M , w � belong to the language = {� M ′ , w ′ � | M ′ is a TM and M ′ accepts w ′ } ? Is D a decidable language? def A TM Facts: L connected is decidable, while A TM is undecidable! Algorithmically solvable problems ≡ decidable languages. Lecture 2 Computability and Complexity 13/15 Lecture 2 Computability and Complexity 14/15 Exam Questions Equivalence of k -tape TM with 1-tape TM. Nondeterministic TM, definition, acceptance of a string, equivalence with ordinary TM. Enumerators, definition, equivalence with ordinary TM. Dovetailing technique. Church-Turing Thesis. Lecture 2 Computability and Complexity 15/15